Find $\sin^5 x + \cos^5 x$ 
Given $$\cos \left[\sqrt{\left(\sin x + \cos x\right)\left(1 - \sin x \cos x \right)}\right] \\{}= \sqrt{\cos \left(\sin x + \cos x \right) \cos \left(1 - \sin x \cos x\right)}.$$
Find $\sin^5 x + \cos^5 x.$

 My try:
\begin{align*}
\sin^5x+\cos^5x
&=(\sin x+\cos x)(\sin^4x-\sin^3x\cos x+\sin^2x\cos^2x-\sin x \cos^3x+\sin^4x)\\
&=(\sin x+\cos x)[(\sin^2x+\cos^2x)^2-(\sin^2x+\cos^2x)\sin x \cos x-\sin^2x\cos^2x]\\
&=(\sin x+\cos x)[1-\sin x\cos x-\sin^2x\cos^2x]\\
&=\frac{1}{4}(\sin x+\cos x)(4-2\sin 2x- \sin^2 2x)\\
\end{align*} I don't know what next?
 A: This is my try.. I'm pretty sure that there are shorter methods to solve this problem. And it's not complete as I have ignored some tedious cases.

Let
$$A=\sin x + \cos x=\sqrt{2}\cos(\pi/4-x)\tag{1}$$
$$B=1-\sin x\cos x=1-\frac{\sin 2x}{2}=1-\frac{\cos[2(\pi/4-x)]}{2}=\frac{3}{2}-\cos^2(\pi/4-x) \tag{2}$$
Multiplying $(1)$ and $(2)$
$$AB=\sin^3{x} + \cos^3{x}=\frac{3\sqrt{2}}{2}\cos(\pi/4-x)-\sqrt{2}\cos^3(\pi/4-x)$$

Given,
$$\cos(\sqrt{AB})=\sqrt{\cos A\cos B}$$
$$\implies2\cos^2(\sqrt{AB})=\cos(A+B)+\cos(A-B)$$
$$2\cos^2(\sqrt{AB})-1=\cos(A+B)+\cos(A-B)-1$$
$$\implies \cos(2\sqrt{AB})=\cos(A+B)+\cos(A-B)-1$$
$$\cos(2\sqrt{AB})-\cos(A-B)=\cos(A+B)-\cos(0)$$
$$-2\sin\bigg(\frac{2\sqrt{AB}+A-B}{2}\bigg)\sin\bigg(\frac{2\sqrt{AB}-A+B}{2}\bigg)=-2\sin^2\bigg(\frac{A+B}{2}\bigg)$$
$$\implies\sin\bigg(\frac{2\sqrt{AB}+A-B}{2}\bigg)\sin\bigg(\frac{2\sqrt{AB}-A+B}{2}\bigg)=
\sin\bigg(\frac{A+B}{2}\bigg)\sin\bigg(\frac{A+B}{2}\bigg)\tag{$\star$}$$
$$\implies\sin\bigg(\frac{2\sqrt{AB}+A-B}{2}\bigg)=\sin\bigg(\frac{2\sqrt{AB}-A+B}{2}\bigg)$$
$$\implies\sin\bigg(\frac{2\sqrt{AB}+A-B}{2}\bigg)-\sin\bigg(\frac{2\sqrt{AB}-A+B}{2}\bigg)=0$$
$$\implies2\cos\big(2\sqrt{AB}\big)\sin\big(A-B\big)=0$$
Note that, for $(\star)$,the graph looks like this.I have only taken the simplest case.

For the sake of simplicity, we will consider $A=B$ as of now and ignore the other cases..
We have,
$$\sqrt{2}\cos(\pi/4-x)=\frac{3}{2}-\cos^2(\pi/4-x)$$
Let $\cos(\pi/4-x)=k$
$$\implies k^2 +\sqrt{2}k-\frac{3}{2}=0$$
$$k=\frac{-3\sqrt{2}}{2},k=\frac{1}{\sqrt{2}}$$
$$\cos(\pi/4-x)=\cos(\pi/4)$$
$$\implies x=2n\pi +\frac{\pi}{2}\text{ or } x=2n\pi\text{ (where }n\in\mathbb{Z})$$
Hence $\sin^5x+\cos^5x=1$ provided that the condition given in the question is true
A: I wanted to develop the idea expressed in comments that LHS is bounded below by $\cos(1)$ while RHS is bounded above by $\cos(1)$ and that we get equality for $x=0\text{ or }\frac{\pi}2$ only.

*

*The LHS part and equality case is not too complicated to prove:

$\require{cancel}\begin{align}f(x)
&=\Big(\cos(x)+\sin(x)\Big)\Big(\overbrace{1}^{\cos(x)^2+\sin(x)^2}-\sin(x)\cos(x)\Big)\\
&=\cos(x)^3+\cancel{\cos(x)\sin(x)^2}-\cancel{\sin(x)\cos(x)^2}+\cancel{\sin(x)\cos(x)^2}+\sin(x)^3-\cancel{\sin(x)^2\cos(x)}\\
&=\cos(x)^3+\sin(x)^3\end{align}$
But notice that $\forall n\in\mathbb N$ since $|\sin(x)|\le 1\implies |\sin(x)|^n\le |\sin(x)|^2$ and same for cosine we get,
$\Big|f(x)\Big|=\Big|\sin(x)^3+\cos(x)^3\Big|\le \Big|\sin(x)\Big|^3+\Big|\cos(x)\Big|^3\le \sin(x)^2+\cos(x)^2=1$
Notice also that $|f(x)|$ has period $\pi$ and that $\cos\searrow$ on $[0,\pi]$
Therefore (and since RHS undefined where $f(x)$ is negative):
$$\cos\left(\sqrt{|f(x)|}\right)\ge \cos(1)\quad\text{and}\quad LHS\ge \cos(1)$$
The equality case is for extremal points.
So let search for critical points $f'(x)=3\sin(x)\cos(x)(\sin(x)-\cos(x))=0\iff x=0,\frac{\pi}2,\frac{\pi}4$ over the $[0,\pi]$ interval, among which only $0,\frac{\pi}2$ lead to $f(x)=1$.


*

*However, I faced a wall when I tried to prove the RHS part, so it seems basilik's approach despite looking more complicated at first sight, is eventually more effective.

Does anyone attempted this approach too ?
